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Hello. In

this video, I am going to show you how you can use the applet i-Cremona

to proceed to simple calculations of a

structure which you are designing.

In the previous video, we have

seen this covering of a public square, temporary

covering, which gives a good idea of what I want to design.

For me, it will be a definitive structure, with an

accessible height of 4 meters at least,

a width of 15 meters and a

total length of 30 meters.

This is a square which is significantly

larger that the one of Bussigny, which I shown you before.

I shall assume, but it is based on reflections,

which correspond to the reality, that the design load qd,

is equal to 1.5 kiloNewtons per square

meter of membrane.

This value takes into account the fact that the membrane

is inclined and that its weight is larger at

some places than at others, but with this value 1.5, I

include these parameters, at least in the case of a conceptual design.

Here is the idea that I have had to cover this place, it is to use

a membrane, with a main cable here, that we can see longitudinally,

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like that, and a membrane which leans on this

cable, and which is anchored on both sides.

So, the membrane, that is what we can see here, with obviously edge

cables, we are not going to go into détails, but

it is important that you should appreciate that it is, here, the

transversal cross-section, and here it is the

longitudinal cross-section.

You can see in dashed lines the dimensions

of the surface to cover, which corresponds to what I have

told you, then 15 by 4 meters height. The first thing that we

have to do is to upload the picture on the i-Structures server.

This functionality is available if we start the applet i-Cremona,

from the usual menu bar of the structures course.

It is not available if we start i-Cremona from an exercise.

We are going to click here, on the bottom, on the button "Add/manage a

picture", and we are going to select the picture

that I have scanned, you can proceed taking a photo

of a drawing, or taking a photo of a structure, if

you wish to understand how a given structure works.

I upload the file with the transversal cross-section to the server.

There we go, it is already done, and I open

the applet, with the background picture that I have defined.

The right part of the applet,

I will not need it this time, therefore I close it, and I use

the zoom functions to center, and enlarge the picture in such

a way that I can comfortably work.

Here, I am going to have a support, here, I am going to have a

second support, between these two supports, a distance of 15 meters.

I want to make the calculation for

a band of membrane which is perpendicular to the screen, of 1

meter width, thus the load which acts on this

band of membrane, is equal to 15 times 1.5 kiloNewtons,

per square meter, times 1 meter of width, that is to say 22.5 kiloNewtons.

I could introduce 15 loads of 1.5 kiloNewton,

but I prefer to introduce 10 loads of 2.25 kiloNewtons.

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1, 2, 3

4, 5 loads on the left, 1, 2, 3, 4, 5 loads on the right.

I activate the funicular polygon and I adjust the initial tangent on the left.

There we go, the structure that I obtain is not correct since

it is blue, it means that it is in compression.

What happens ?

Well, I forgot something, it is that in the middle of this

membrane, acts the load-bearing cable which pulls all the structure upwards.

So, I am going to introduce a force upwards,

quite large, I am going to put 10 kiloNewtons.

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There we go, like that.

And this force does not pull enough

upwards.

To be able to work well, I need to

zoom in, to put to scale, the forces.

Then I push this button, scaling

of the forces, and I press on shift to get

the opposite effect and then I am going to get a

zoom-out on the forces, like that it is okay.

I unlock the scaling button, but I keep the shift

button which enables me to increase this force, without changing its orientation.

There I go, it goes down more and more, and

suddenly, at once, the structure reappears in red,

in the upper part, until it can come

here, to be precisely superimposed on the structure that I had sketched.

The required force is roughly 32 kiloNewtons.

We are going to remember of this value,

and we are now going to look at

the longitudinal cross-section. We come back to the list of our

pictures, we are going to search the picture of the longitudinal cross-section.

We upload it on the server, and we open it in the applet.

Once again, we do not need the right part.

So, we are going to use the zoom-in function to get

a little bit of space.

We are going to have a support on the left, here,

another support on the right, and then, on a distance of roughly 30 meters,

we will have 32 kiloNewtons per meter, which is the load

which comes from each meter of membrane in the transversal direction.

32 times 30, it is equal to 960 kiloNewtons, I am going to introduce

10 loads of 96 kiloNewtons, along this cable.

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1, 2, 3, 4, 5, 6,

7, 8, 9 and 10 loads. I activate the funicular

polygon and I give to the cable the same shape that I had envisioned.

We can see that there are little imprecision, I replace the support more

correctly, and I adjust one more time the geometry, and then

I reach a geometry which is quite close to the one

I had sketched, it was a free-hand drawing,

with an internal force of roughly 2350 on the left, 2390,

let's say 2400 kiloNewtons, which acts in this structure.

It is a quite significant internal force, if you make a dimensioning

calculation to get the required diameter of the cable, consider 300

Newtons per square millimeters as strength, and you will see

that the required diameter of the cable is quite large.

I think that in this case, it will

probably be necessary to use more than one cable, as we

have incidentally seen it in other configurations of membranes,

because, probably, the internal forces will be too concentrated.

On the basis of what we have now, we use a

free-body, we know the internal force here, we can obtain

the internal force that I am only going to draw with a line,

the tensile internal force in this cable, the compressive internal force in this

compressed element, likewise, here on the left.

And thus we can quickly obtain all

the dimensions, diameters of the posts, diameters of the cables,

to have an idea of the amount of materials, and

it will enable us to quickly qualify, or to disqualify a structure.

In this video, we have seen how you can introduce yourself

a picture to be used as a basis for the calculations in i-Cremona.

You have also seen how to quickly get an

idea of the internal forces which act on a structure,

to be able to dimension it.